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3 years ago

Round48.073 to the nearest hundredth

Mathematics

2 answers:

Lelu [443]3 years ago

8 0

Answer:

48.07

Step-by-step explanation:

Gnesinka [82]3 years ago

4 0

The answer is: 48.07

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Step-by-step explanation:

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Answer:

5/162

Step-by-step explanation:

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2 years ago

120 - 400 = 520 Re-Write as an Addition problem

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Step-by-step explanation:

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2 years ago

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18x39 how do u do Distribute property

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2 years ago

Find all values of in the interval [0, 2π] that satisfy the equation. Enter your answers from smallest to largest. (Enter NONE i

Mrrafil [7]

Answer:

Step-by-step explanation:

Given the equation $2\sin^2(x) = 1 \\$ , we are to find all the values of x in the interval [0, 2π] that satisfies the equation.

Simplifying the equation:

$2\sin^2(x) = 1 \\\\sin^2x = \frac{1}{2} \\\\$

Square both sides of the equation;

$\sqrt{sin^2x } = \sqrt{\frac{1}{2} } \\\\sin(x) = \sqrt{0.5}\\ \\sin(x) = 0.7071\\\\x = sin^{-1}0.7071\\\\x \approx 45^0$

The general formula for sin(x) is x = nπ + (-1)ⁿ∝ where n = 0, 1, 2, 3...

Since ∝ = 45° = π/4.

x = nπ + (-1)ⁿπ/4

when n = 1

x = π + (-1)¹π/4

x = π - π/4

x = 3π/4

<em>Hence the values of x within the given interval are π/4 and 3π/4</em>

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3 years ago